foj 2111 Problem 2111 Min Number

Problem 2111 Min Number

Accept: 1025    Submit: 2022
Time Limit: 1000 mSec    Memory Limit : 32768 KB

Problem Description

Now you are given one non-negative integer n in 10-base notation, it will only contain digits ('0'-'9'). You are allowed to choose 2 integers i and j, such that: i!=j, 1≤i<j≤|n|, here |n| means the length of n’s 10-base notation. Then we can swap n[i] and n[j].

For example, n=9012, we choose i=1, j=3, then we swap n[1] and n[3], then we get 1092, which is smaller than the original n.

Now you are allowed to operate at most M times, so what is the smallest number you can get after the operation(s)?

Please note that in this problem, leading zero is not allowed!

Input

The first line of the input contains an integer T (T≤100), indicating the number of test cases.

Then T cases, for any case, only 2 integers n and M (0≤n<10^1000, 0≤M≤100) in a single line.

Output

For each test case, output the minimum number we can get after no more than M operations.

Sample Input

3 9012 0 9012 1 9012 2

Sample Output

9012 1092 1029 

 

思路：贪心模拟，头部尽量与尾部的最小值交换。

AC代码：

#include <iostream>
#include<algorithm>
#include<vector>
#include<cstring>
#include<queue>
#include<string>
using namespace std;
#define N_MAX 10000+10
#define V_MAX 1000+10
#define INF 0x3f3f3f3f
int m;
string n;
void change(string &s,int n){//当前调整s的第n位
    char c='9'+1;int id;
  for(int i=s.size()-1;i>n;i--){
     if(c>s[i]){
        if(n==0&&s[i]=='0')continue;
        c=s[i];id=i;
      }
  }
  if(c<s[n])swap(s[n],s[id]);
}
int main(){
    int t;scanf("%d",&t);
    while(t--){
       cin>>n>>m;
       int cnt=0;
       string s;
       while(m){
            s=n;
       while(s==n&&cnt<s.size()-1){
        change(n,cnt);
          cnt++;
       }
       if(cnt>=s.size()-2)break;//已经不需要交换了
       m--;
       }
       cout<<n<<endl;
    }
    return 0;
}